(1)∵AB=AC,BC=12,AF⊥BC于点F, ∴BF=FC=6. ∵⊙O经过点F,并分别与AB、AC边切于点D、E. ∴BD=BF=6,CE=CF=6. ∵AB=AC=10, ∴AD=AE=4,∴AD:AB=AE:AC,∴DE∥BC, ∴DE:BC=AD:AB,即DE:12=4:10,∴DE=4.8, ∴△ADE的周长=AD+DE+AE=4+4+4.8=12.8.
(2)∵AF⊥BC于点F,∴∠AFB=90°. ∵AB=10,BF=6,∴AF==8. ∵⊙O与AC边切于点D,∴∠ADO=90°. ∴∠ADO=∠AFB,且OD=OF. ∵∠OAD=∠BAF,∴△ADO∽△AFB, ∴AO:AB=OD:BF, 即(8-OD):10=OD:6,∴OD=3, ∴S⊙O=π•OD2=9π. |